Question

25.0 mL of 0.138M HCl were used to completely neutralize 39.0 mL of KOH. What is the concentration of the KOH used in this lab?

Answer 1

Answer:

0.0885M

Explanations:

In order to determine the concentration of the KOH used in this lab, we will use the dilution formula expressed as

$C_1V_1=C_2V_2$

where:

• C₁ and C₂ are the ,initial and final concentrations

,

• V₁ and V₂ are the ,initial and final volumes

Given the following parameters

• C₁ = 0.138M

,

• V₁ = 25.0mL = 0.025L

,

• V₂ = 39.0mL = 0.039L

Required

Final concentration C₂

Substitute the given parameter into the formula to have:

$\begin{gathered} C_2=\frac{C_1V_1}{V_2} \\ C_2=\frac{0.138\times0.025}{0.039} \\ C_2=\frac{0.00345}{0.039} \\ C_2=0.0885M \end{gathered}$

Hence the concentration of the KOH used in this lab is 0.0885M